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For how many integers n between 5 and 20, inclusive, is the
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03 May 2019, 02:05
03 May 2019, 02:05
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Question Stats:
89% (01:18) correct
10% (00:42) wrong based on 37 sessions
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For how many integers n between 5 and 20, inclusive, is the sum of \(3n\), \(9n\), and \(11n\) greater than 200?
(A) 4
(B) 8
(C) 12
(D) 16
(E) 20
(A) 4
(B) 8
(C) 12
(D) 16
(E) 20
Re: For how many integers n between 5 and 20, inclusive, is the
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04 Dec 2019, 12:28
04 Dec 2019, 12:28
1
The easiest way to solve this problem is noticing that:
for n=8 3n + 9n + 11n = 184
for n=9 3n + 9n + 11n = 207
Therefore all the numbers between 9 and 20, inclusive, will respect our condition.
The right answer is C.
for n=8 3n + 9n + 11n = 184
for n=9 3n + 9n + 11n = 207
Therefore all the numbers between 9 and 20, inclusive, will respect our condition.
The right answer is C.
Re: For how many integers n between 5 and 20, inclusive, is the
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05 Dec 2019, 01:51
05 Dec 2019, 01:51
1
3n+9n+11n > 200
23n > 200
n>8.69 Given range of values for n is 5 <= n <= 20, So from 9 to 20 that is 12 values.
23n > 200
n>8.69 Given range of values for n is 5 <= n <= 20, So from 9 to 20 that is 12 values.
